To determine To calculate: The resultant internal loadings acting in the beam at cross sections through points D and E . Explanation Given information: The length of the beam is 20-ft. Calculation: Sketch the Free Body Diagram of Member AB as shown in Figure 1. Refer to Figure 1. Find the loading intensity as shown below. P A B = 1 2 × Intensity of loading × Length of loading Substitute 1.5 kip / ft for intensity of loading and 12 ft for length of loading. P A B = 1 2 × 1.5 × 12 = 9 kip Find the support reactions at A and B as shown below. Take moment about B is Equal to zero. ∑ M B = 0 9 × 4 − A y × 12 = 0 36 − 12 A y = 0 A y = 3 kip Summation of forces along horizontal direction is Equal to zero. ∑ F x = 0 B x = 0 Summation of forces along vertical direction is Equal to zero. ∑ F y = 0 B y + 3 − 9 = 0 B y − 6 = 0 B y = 6 kip Sketch the Free Body Diagram of segment AD as shown in Figure 2. Refer to Figure 2. Find the internal loadings as shown below. Apply the Equations of Equilibrium as shown below. Summation of forces along horizontal direction is Equal to zero. ∑ F x = 0 N D = 0 Summation of forces along vertical direction is Equal to zero. ∑ F y = 0 3 − 2.25 − V D = 0 0.75 − V D = 0 V D = 0.75 kip Take moment about D is Equal to zero

9th Edition

ISBN: 9780133254426

Author: Russell C. Hibbeler

Publisher: Prentice Hall

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Types of loading

In mechanical terminology, the term load indicates that external force/weight, either constant or variable, acts on a mechanical member/structure. The weight of an object of a specific material is also an external load, and the external force that works …

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Chapter 1.2, Problem 1.5P

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To calculate: The resultant internal loadings acting in the beam at cross sections through points D and E.

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Chapter 1.2, Problem 1.4P

Chapter 1.2, Problem 1.6P

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